得到每個類別過濾器的最後一個記錄日期

Question

我有這個表SQL Fiddle

項目表：

+----+----------+ 
| id | name | 
+----+----------+ 
| 1 | Facebook | 
| 2 | Twitter | 
| 3 | Amazon | 
+----+----------+ 

價格表：

+----+-----------+---------+-----------------------------+ 
| id | buy | item_id |    created_at  | 
+----+-----------+---------+-----------------------------+ 
| 1 | 43000 | 1 | June, 18 2014 17:31:04+0000 | 
| 2 | 44000 | 1 | June, 19 2014 17:31:04+0000 | 
| 3 | 30000 | 2 | June, 20 2014 17:31:04+0000 | 
| 4 | 33000 | 2 | June, 21 2014 17:31:04+0000 | 
| 5 | 20000 | 3 | June, 22 2014 17:31:04+0000 | 
| 6 | 21000 | 3 | June, 23 2014 17:31:04+0000 | 
+----+-----------+---------+-----------------------------+ 

我想每個項目和一個最後的價格在基於價格日期的最後價格買入區域之前

期望的輸出：

+----+---------+-----------------+---------+ 
| id | buy | last_before_buy | item_id | 
+----+---------+-----------------+---------+ 
| 10 | 45000 |  43000  | 3 | 
| 7 | 33000 |  31000  | 2 | 
| 4 | 23000 |  23000  | 1 | 
+----+---------+-----------------+---------+ 

Answer 1

這裏是另一種方式來做到這一點：

select a.id, a.buy, b.buy last_before_buy, a.item_id 
from (select * from prices WHERE (created_at <= NOW() - INTERVAL 5 DAY) order by id desc) a 
join (select * from prices order by id desc) b on a.item_id = b.item_id and a.id > b.id 
group by a.item_id; 

fiddle

Answer 2

您可以用substring_index()/group_concat()的方法來做到這點：

select max(id) as id, 
     substring_index(group_concat(buy order by created_at desc), ',', 1) as buy, 
     substring_index(substring_index(group_concat(buy order by created_at desc), ',', 2), ',', -1) as lastbuy, 
     item_id 
from prices p 
group by item_id;